题目描述
Farmer John has an old-time thresher (wheat harvester) that requires belts to be installed on various gears to turn the parts. The engine drives pulley 1 in a clockwise direction which attaches via a belt to pulley 2. Pulley 2 attaches via a belt to pulley 3 and so on through a total of N (2 <= N <= 1,000) pulleys (and N-1 belts).
The diagram above depicts the two ways a belt can be installed between two gears. In this illustration, pulley 1's belt directly drives pulley 2 (a 'straight' connection) and thus they will rotate in the same direction. Pulley 3 drives pulley 4 via a 'crossed belt' that reverses the direction of the rotation.
Given a list of the belt types that connect the pulleys along with the fact that pulley 1 is driven in a clockwise direction by the engine, determine the drive direction of pulley N. Each belt is described by three integers:
* S_i -- the driving (source) pulley
* D_i -- the driven (destination) pulley
* C_i -- the connection type (0=straight, 1=crossed)
Unfortunately, FJ lists the belts in random order.
By way of example, consider the illustration below. N = 4, and pulley 1 is driven clockwise by the thresher engine. Straight
belts drive pulley 2 and then pulley 3, so they rotate clockwise. The crosswise belt reverses the rotation direction so pulley 4 (pulley N) rotates counterclockwise. POINTS: 70 约翰有一个过时的收割机，需要在它的各种滑轮上装配皮带才能让收割机的各个部分运作起 来.引擎能够驱动滑轮1向顺时针方向转动，滑轮1通过一条皮带又连接到滑轮2.滑轮2又通过一 条皮带连接到滑轮3，等等，总共有N(2 <= N <= 1000)个滑轮和N - 1条皮带.
皮带连接两个滑轮有两种方式：直接连接和交叉连接.直接连接的两个滑轮旋转方向相同， 即同为顺时针或同为逆时针.交叉连接的两个滑轮旋转方向相反.
现在给出一个列表，里面列出所有皮带的连接方式.已经知道滑轮1被引擎驱动着向顺时针方 向转动.每一条皮带由下面三个数定义：
•驱动滑轮S，输入驱动力的滑轮.
•被驱动滑轮D;，被驱使转动的滑轮.
•连接类型C，0表示直接连接，1表示交叉连接.
不幸的是，约翰的这个列表中，皮带的顺序是混乱的.所以请你写一个程序来求出滑轮N的 转动方向.
输入输出格式
输入格式：
-
Line 1: A single integer: N
- Lines 2..N: Each line describes a belt with three integers: S_i, D_i, and C_i
输出格式：
- Line 1: A single integer that is the rotation direction for pulley N (0=clockwise, 1=counterclockwise)
输入输出样例
说明
As in the example illustration.
思路：搜索。
#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define MAXN 1010 using namespace std; int n,tot; int vis[MAXN]; int to[MAXN],head[MAXN],net[MAXN],cap[MAXN]; void add(int u,int v,int w){ to[++tot]=v;cap[tot]=w;net[tot]=head[u];head[u]=tot; } void dfs(int now,int fa){ for(int i=head[now];i;i=net[i]) if(to[i]!=fa){ if(cap[i]==0) vis[to[i]]=vis[now]; else vis[to[i]]=!vis[now]; dfs(to[i],now); } } int main(){ scanf("%d",&n); for(int i=1;i<n;i++){ int x,y,z; scanf("%d%d%d",&x,&y,&z); add(x,y,z); } vis[1]=0; dfs(1,0); cout<<vis[n]; }